If (a – b) = 3, (b – c) = 5 and (c – a) = 1, then the value of $frac{a^{3}+b^{3}+c^{3}-3 a b c}{a+b+c}$ is

a³ + b³ + c³ – 3abc = (a + b + c)

(a² + b² +c² – ab – bc – ac)

$=frac{1}{2}(a+b+c)left(2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 a c ight)$

$=frac{1}{2}(a+b+c)left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2} ight]$

$herefore quad frac{a^{3}+b^{3}+c^{3}-3 a b c}{a+b+c}$

$=frac{1}{2}left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2} ight]$

$=frac{1}{2}(9+25+1)=frac{35}{2}=17.5$